Show that $x=2\ln(3x-2)$ can be written as $x=\frac{1}{3}(e^{x/2}+2)$

Question

Show that $x=2\ln(3x-2)$ can be written as $x=\dfrac{1}{3}(e^{x/2}+2)$.

Is there a rule for this?

Answer 1

Solve for the "other" $x$. Notice that: $$ \begin{align*} x &= 2\ln(3x-2) \\ \dfrac{x}{2} &= \ln(3x-2) \\ e^{x/2} &= 3x-2 \\ e^{x/2}+2 &= 3x \\ \dfrac{1}{3}(e^{x/2}+2) &= x \\ \end {align*} $$ as desired.

Answer 2

exponential function is injective, so $x = y$ iff $e^x = e^y$. use this fact and then some elementary symbol manipulations should give you the desired result (if it's true of course)

Answer 3

It’s just the relationship between logs and exponents. Starting with $x=2\ln(3x-2)$, divide both sides by $2$ to get

$$\frac{x}2=\ln(3x-2)\;.$$

Now exponentiate on both sides:

$$e^{\frac{x}2}=e^{\ln(3x-2)}=3x-2\;,$$

because by definition $e^{\ln u}=u$ for all $u>0$. Finally, add $2$ to both sides to get

$$e^{\frac{x}2}+2=3x\;,$$

and multiply both sides by $\frac13$, and you’re done.